Let `n = 2006!` Then `(1)/(log_(2)n) + (1)/(log_(3)n) + ... + (1)/(log_(2006)n) =`

Options

A. 2006
B. 2005
C. 2005!
D. 1

Answer

To solve the problem, we need to evaluate the expression:

\[ \frac{1}{\log_2 n} + \frac{1}{\log_3 n} + \cdots + \frac{1}{\log_{2006} n} \]

where \( n = 2006! \).

### Step 1: Use the Change of Base Formula
We start by applying the change of base formula for logarithms, which states that:

\[ \log_a b = \frac{1}{\log_b a} \]

Thus, we can rewrite each term in the expression:

\[ \frac{1}{\log_k n} = \log_n k \]

So, we can rewrite the entire sum as:

\[ \log_n 2 + \log_n 3 + \cdots + \log_n 2006 \]

### Step 2: Combine the Logarithms
Using the property of logarithms that states \( \log_a b + \log_a c = \log_a (bc) \), we can combine the logarithms:

\[ \log_n (2 \times 3 \times 4 \times \cdots \times 2006) \]

This product can be recognized as \( \log_n (2006!) \).

### Step 3: Substitute \( n \)
Since we know \( n = 2006! \), we can substitute this into our expression:

\[ \log_{2006!} (2006!) \]

### Step 4: Evaluate the Logarithm
Using the property of logarithms that states \( \log_a a = 1 \), we find:

\[ \log_{2006!} (2006!) = 1 \]

### Final Answer
Thus, the value of the original expression is:

\[ \frac{1}{\log_2 n} + \frac{1}{\log_3 n} + \cdots + \frac{1}{\log_{2006} n} = 1 \]